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CHAPTER 14 SECTION 1: ANALYSIS OF VARIANCE
MULTIPLE CHOICE
1. In one-way ANOVA, the amount of total variation that is unexplained is measured by the:
a. sum of squares for treatments.
b. sum of squares for error.
c. total sum of squares.
d. degrees of freedom.
ANS: B
PTS: 1
REF: SECTION 14.1
2. The test statistic of the single-factor ANOVA equals:
a. sum of squares for treatments / sum of squares for error.
b. sum of squares for error / sum of squares for treatments.
c. mean square for treatments / mean square for error.
d. mean square for error / mean square for treatments.
ANS: C
PTS: 1
REF: SECTION 14.1
3. In a single-factor analysis of variance, MST is the mean square for treatments and MSE is the mean
square for error. The null hypothesis of equal population means is rejected if:
a. MST is much smaller than MSE.
b. MST is much larger than MSE.
c. MST is equal to MSE.
d. None of these choices.
ANS: B
PTS: 1
REF: SECTION 14.1
4. Which of the following is not a required condition for one-way ANOVA?
a. The sample sizes must be equal.
b. The populations must all be normally distributed.
c. The population variances must be equal.
d. The samples for each treatment must be selected randomly and independently.
ANS: A
PTS: 1
REF: SECTION 14.1
5. The analysis of variance is a procedure that allows statisticians to compare two or more population:
a. means.
b. proportions.
c. variances.
d. standard deviations.
ANS: A
PTS: 1
REF: SECTION 14.1
6. The distribution of the test statistic for analysis of variance is the:
a. normal distribution.
b. Student t-distribution.
c. F-distribution.
d. None of these choices.
ANS: C
PTS: 1
REF: SECTION 14.1
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copied, or distributed without the prior consent of the publisher.
7. In a one-way ANOVA, error variability is computed as the sum of the squared errors, SSE, for all
values of the response variable. This variability is the:
a. the total variation.
b. within-treatments variation.
c. between-treatments variation.
d. None of these choices.
ANS: B
PTS: 1
REF: SECTION 14.1
8. In the one-way ANOVA where there are k treatments and n observations, the degrees of freedom for
the F-statistic are equal to, respectively:
a. n and k.
b. k and n.
c. n  k and k  1.
d. k  1 and n  k.
ANS: D
PTS: 1
REF: SECTION 14.1
9. In the one-way ANOVA where k is the number of treatments and n is the number of observations in all
samples, the degrees of freedom for treatments is given by:
a. k  1
b. n  k
c. n  1
d. n  k + 1
ANS: A
PTS: 1
REF: SECTION 14.1
10. In ANOVA, the F-test is the ratio of two sample variances. In the one-way ANOVA (completely
randomized design), the variance used as a numerator of the ratio is:
a. mean square for treatments.
b. mean square for error.
c. total sum of squares.
d. None of these choices.
ANS: A
PTS: 1
REF: SECTION 14.1
11. In a completely randomized design for ANOVA, the numerator and denominator degrees of freedom
are 4 and 25, respectively. The total number of observations must equal:
a. 24
b. 25
c. 29
d. 30
ANS: D
PTS: 1
REF: SECTION 14.1
12. The number of degrees of freedom for the denominator in one-way ANOVA test involving 4
population means with 15 observations sampled from each population is:
a. 60
b. 19
c. 56
d. 45
ANS: C
PTS: 1
REF: SECTION 14.1
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13. The value of the test statistic in a completely randomized design for ANOVA is F = 6.29. The degrees
of freedom for the numerator and denominator are 5 and 10, respectively. Using an F table, the most
accurate statements to be made about the p-value is that it is:
a. greater than 0.05
b. between 0.025 and 0.050.
c. between 0.010 and 0.025.
d. between 0.001 and 0.010.
ANS: D
PTS: 1
REF: SECTION 14.1
14. In one-way ANOVA, the term refers to the:
a. sum of the sample means.
b. sum of the sample means divided by the total number of observations.
c. sum of the population means.
d. weighted average of the sample means.
ANS: D
PTS: 1
REF: SECTION 14.1
15. For which of the following is not a required condition for ANOVA?
a. The populations are normally distributed.
b. The population variances are equal.
c. The samples are independent.
d. All of these choices are required conditions for ANOVA.
ANS: D
PTS: 1
REF: SECTION 14.1
16. One-way ANOVA is applied to independent samples taken from three normally distributed
populations with equal variances. Which of the following is the null hypothesis for this procedure?
a. 1 + 2 + 3 = 0
b. 1 + 2 + 3  0
c. 1 = 2 = 3 = 0
d. 1 = 2 = 3
ANS: D
PTS: 1
REF: SECTION 14.1
17. In the one-way ANOVA where k is the number of treatments and n is the number of observations in all
samples, the number of degrees of freedom for error is:
a. k  1
b. n  k
c. n  1
d. n  k + 1
ANS: B
PTS: 1
REF: SECTION 14.1
18. How does conducting multiple t-tests compare to conducting a single F-test?
a. Multiple t-tests increases the chance of a Type I error.
b. Multiple t-tests decreases the chance of a Type I error.
c. Multiple t-tests does not affect the chance of a Type I error.
d. This comparison cannot be made without knowing the number of populations.
ANS: A
PTS: 1
REF: SECTION 14.1
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19. In one-way analysis of variance, between-treatments variation is measured by the:
a. SSE
b. SST
c. SS(Total)
d. standard deviation
ANS: B
PTS: 1
REF: SECTION 14.1
20. One-way ANOVA is applied to independent samples taken from four normally distributed populations
with equal variances. If the null hypothesis is rejected, then we can infer that
a. all population means are equal.
b. all population means differ.
c. at least two population means are equal.
d. at least two population means differ.
ANS: D
PTS: 1
REF: SECTION 14.1
21. Consider the following partial ANOVA table:
Source of Variation
Treatments
Error
Total
SS
75
60
135
df
*
*
19
MS
25
3.75
F
6.67
The numerator and denominator degrees of freedom for the F-test (identified by asterisks) are
a. 4 and 15
b. 3 and 16
c. 15 and 4
d. 16 and 3
ANS: B
PTS: 1
REF: SECTION 14.1
22. Consider the following ANOVA table:
Source of Variation
Treatments
Error
Total
SS
4
30
34
df
2
12
14
MS
2.0
2.5
F
0.80
The number of treatments is
a. 13
b. 5
c. 3
d. 12
ANS: C
PTS: 1
REF: SECTION 14.1
23. In one-way analysis of variance, within-treatments variation is measured by:
a. sum of squares for error.
b. sum of squares for treatments.
c. total sum of squares.
d. standard deviation.
ANS: A
PTS: 1
REF: SECTION 14.1
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24. Consider the following ANOVA table:
Source of Variation
Treatments
Error
Total
SS
128
270
398
df
4
25
29
MS
32
10.8
F
2.963
The total number of observations is:
a. 25
b. 29
c. 30
d. 32
ANS: C
PTS: 1
REF: SECTION 14.1
25. In one-way analysis of variance, if all the sample means are equal, then the:
a. total sum of squares is zero.
b. sum of squares for error is zero.
c. sum of squares for treatments is zero.
d. sum of squares for error equals sum of squares for treatments.
ANS: C
PTS: 1
REF: SECTION 14.1
26. Which of the following components in an ANOVA table is not additive?
a. Sum of squares
b. Degrees of freedom
c. Mean squares
d. All of these choices are additive.
ANS: C
PTS: 1
REF: SECTION 14.1
27. In which case can an F-test be used to compare two population means?
a. For one tail tests only.
b. For two tail tests only.
c. For either one or two tail tests.
d. None of these choices.
ANS: B
PTS: 1
REF: SECTION 14.1
28. The F-test statistic in a one-way ANOVA is equal to:
a. MST/MSE
b. SST/SSE
c. MSE/MST
d. SSE/SST
ANS: A
PTS: 1
REF: SECTION 14.1
29. The numerator and denominator degrees of freedom for the F-test in a one-way ANOVA are,
respectively,
a. (n  k) and (k  1)
b. (k  1) and (n  k)
c. (k  n) and (n  1)
d. (n  1) and (k  n)
ANS: B
PTS: 1
REF: SECTION 14.1
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30. Which of the following statements is false?
a. F = t2
b. The F-test can be used instead of a two tail t-test when you compare two population
means.
c. Doing three t-tests is statistically equivalent to doing one F-test when you compare three
population means.
d. All of these choices are true.
ANS: C
PTS: 1
REF: SECTION 14.1
TRUE/FALSE
31. We use the analysis of variance (ANOVA) technique to compare two or more population means.
ANS: T
PTS: 1
REF: SECTION 14.1
32. The sum of squares for treatments, SST, achieves its smallest value (zero) when all the sample means
are equal.
ANS: T
PTS: 1
REF: SECTION 14.1
33. The analysis of variance (ANOVA) technique analyzes the variance of the data to determine whether
differences exist between the population means.
ANS: T
PTS: 1
REF: SECTION 14.1
34. Conducting t-tests for each pair or population means is statistically equivalent to conducting one F-test
comparing all the population means.
ANS: T
PTS: 1
REF: SECTION 14.1
35. The sum of squares for error is also known as the between-treatments variation.
ANS: F
PTS: 1
REF: SECTION 14.1
36. The F-test used in one-way ANOVA is an extension of the t-test of 1  2.
ANS: T
PTS: 1
REF: SECTION 14.1
37. In one-way ANOVA, the total variation SS(Total) is partitioned into two sources of variation: the sum
of squares for treatments (SST) and the sum of squares for error (SSE).
ANS: T
PTS: 1
REF: SECTION 14.1
38. In ANOVA, a criterion by which the populations are classified is called a factor.
ANS: T
PTS: 1
REF: SECTION 14.1
39. We can use the F-test to determine whether 1 is greater than 2.
ANS: F
PTS: 1
REF: SECTION 14.1
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copied, or distributed without the prior consent of the publisher.
40. In one-way ANOVA, the test statistic is defined as the ratio of the mean square for error (MSE) and
the mean square for treatments (MST), namely, F = MSE / MST.
ANS: F
PTS: 1
REF: SECTION 14.1
41. The sum of squares for treatments (SST) is the variation attributed to the differences between the
treatment means, while the sum of squares for error (SSE) measures the within-treatment variation.
ANS: T
PTS: 1
REF: SECTION 14.1
42. The F-test in ANOVA tests whether or not the population variances are equal.
ANS: F
PTS: 1
REF: SECTION 14.1
43. If the numerator (MST) degrees of freedom is 3 and the denominator (MSE) degrees of freedom is 18,
the total number of observations must equal 21.
ANS: F
PTS: 1
REF: SECTION 14.1
44. The F-statistic in a one-way ANOVA represents the variation within the treatments divided by the
variation between the treatments.
ANS: F
PTS: 1
REF: SECTION 14.1
45. The sum of squares for error (SSE) measures the amount of variation that is explained by the ANOVA
model, while the sum of squares for treatments (SST) measures the amount of variation that remains
unexplained.
ANS: F
PTS: 1
REF: SECTION 14.1
46. The distribution of the test statistic for analysis of variance is the F-distribution.
ANS: T
PTS: 1
REF: SECTION 14.1
47. The analysis of variance (ANOVA) tests hypotheses about population variances and requires all the
population means to be equal.
ANS: F
PTS: 1
REF: SECTION 14.1
48. The F-test in ANOVA is an expansion of the t-test for two independent population means.
ANS: T
PTS: 1
REF: SECTION 14.1
49. When the F-test is used for ANOVA, the rejection region is always in the right tail.
ANS: T
PTS: 1
REF: SECTION 14.1
50. The within-treatments variation provides a measure of the amount of variation in the response
variables that is caused by the treatments.
ANS: F
PTS: 1
REF: SECTION 14.1
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COMPLETION
51. The ANOVA procedure tests to determine whether differences exist between two or more population
____________________.
ANS: means
PTS: 1
REF: SECTION 14.1
52. The null hypothesis of ANOVA is that all the population means are ____________________.
ANS: equal
PTS: 1
REF: SECTION 14.1
53. The alternative hypothesis of ANOVA is that ____________________ population means are different.
ANS:
at least 2
at least two
two or more
2 or more
PTS: 1
REF: SECTION 14.1
54. In ANOVA the populations are classified according to one or more criterion, called
____________________.
ANS: factors
PTS: 1
REF: SECTION 14.1
55. SST measures the variation ____________________ treatments.
ANS: between
PTS: 1
REF: SECTION 14.1
56. SSE measures the variation ____________________ treatments.
ANS: within
PTS: 1
REF: SECTION 14.1
57. The F-test statistic in ANOVA is equal to MS____________________ divided by
MS____________________ and H0 is rejected for ____________________ values of F.
ANS: T; E; large
PTS: 1
REF: SECTION 14.1
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58. If SST explains a significant portion of the total variation, we conclude that the population means
____________________ (do/do not) differ.
ANS: do
PTS: 1
REF: SECTION 14.1
59. The F-test in ANOVA requires that the random variable be ____________________ distributed with
equal ____________________.
ANS:
normally; variances
normal; variances
PTS: 1
REF: SECTION 14.1
60. If we square the t-statistic for two means, the result is the ____________________-statistic.
ANS: F
PTS: 1
REF: SECTION 14.1
SHORT ANSWER
TV Viewing Habits
A statistician employed by a television rating service wanted to determine if there were differences in
television viewing habits among three different cities in California. She took a random sample of five
adults in each of the cities and asked each to report the number of hours spent watching television in
the previous week. The results are shown below. (Assume normal distributions with equal variances.)
Hours Spent Watching Television
San Diego
Los Angeles
San Francisco
25
28
23
31
33
18
18
35
21
23
29
17
27
36
15
61. {TV Viewing Habits Narrative} Set up the ANOVA Table. Use  = 0.05 to determine the critical
value.
ANS:
Source of Variation
Treatments
Error
Total
PTS: 1
SS
450.533
184.400
634.933
df
2
12
14
MS
225.267
15.367
F
14.659
P-value
0.0006
F critical
3.885
REF: SECTION 14.1
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copied, or distributed without the prior consent of the publisher.
62. {TV Viewing Habits Narrative} Can she infer at the 5% significance level that differences in hours of
television watching exist among the three cities?
ANS:
H0: 1 = 2 = 3 vs. H1: At least two means differ
Conclusion: Reject the null hypothesis. Yes, differences in mean hours of television watching exist in
at least two of the three cities, according to this data.
PTS: 1
REF: SECTION 14.1
Pain Formulas
A pharmaceutical manufacturer has been researching new formulas to provide quicker relief of minor
pains. Their laboratories have produced three different formulas and they want to determine if the
different formulas produce different responses. Fifteen people who complained of minor pains were
recruited for an experiment; five were randomly assigned to each formula. Each person was asked to
take the medicine and report the length of time until some relief was felt (minutes). The results are
shown below. (Assume normal distributions with equal variances.)
Time in Minutes Until Relief Is Felt (min)
Formula 1
Formula 2
Formula 3
4
2
6
8
5
7
6
3
7
9
7
8
8
1
6
63. {Pain Formulas Narrative} Set up the ANOVA Table. Use  = 0.05 to determine the critical value.
ANS:
Source of Variation
Treatments
Error
Total
PTS: 1
SS
36.4
42.0
78.4
df
2
12
14
MS
18.2
3.5
F
5.2
P-value
0.0236
F critical
3.885
REF: SECTION 14.1
64. {Pain Formulas Narrative} Do these data provide sufficient evidence to indicate that differences in the
average time of relief exist among the three formulas? Use  = 0.05.
ANS:
H0: 1 = 2 = 3 vs. H1: At least two means differ
Conclusion: Reject the null hypothesis. Yes, differences in the mean time of relief exist in at least two
of the three formulas, according to this data.
PTS: 1
REF: SECTION 14.1
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copied, or distributed without the prior consent of the publisher.
Pizza Customers
The marketing manager of a pizza chain is in the process of examining some of the demographic
characteristics of her customers. In particular, she would like to investigate the belief that the ages of
the customers of pizza parlors, hamburger emporiums, and fast-food chicken restaurants are different.
The ages of eight randomly selected customers of each of the restaurants are recorded and listed
below. From previous analyses we know that the ages are normally distributed with equal variances
for each group.
Pizza
23
19
25
17
36
25
28
31
Customers' Ages
Hamburger
26
20
18
35
33
25
19
17
Chicken
25
28
36
23
39
27
38
31
65. {Pizza Customers Narrative} Set up the ANOVA Table. Use  = 0.05 to determine the critical value.
ANS:
Source of Variation
Treatments
Error
Total
PTS: 1
SS
203.583
863.750
1067.333
df
2
21
23
MS
101.792
41.131
F
2.475
P-value
0.1084
F critical
3.467
REF: SECTION 14.1
66. {Pizza Customers Narrative} Do these data provide enough evidence at the 5% significance level to
infer that there are differences in ages among the customers of the three restaurants?
ANS:
H0: 1 = 2 = 3 vs. H1: At least two means differ
Conclusion: Don't reject the null hypothesis. Cannot conclude that average age differences exist
among the customers of the three restaurants, according to this data.
PTS: 1
REF: SECTION 14.1
GMAT Scores
A recent college graduate is in the process of deciding which one of three graduate schools he should
apply to. He decides to judge the quality of the schools on the basis of the Graduate Management
Admission Test (GMAT) scores of those who are accepted into the school. A random sample of six
students in each school produced the following GMAT scores. Assume that the data are normally
distributed with equal variances for each school.
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School 1
650
620
630
580
710
690
GMAT Scores
School 2
105
550
700
630
600
650
School 3
590
510
520
500
490
530
67. {GMAT Scores Narrative} Set up the ANOVA Table. Use  = 0.05 to determine the critical value.
ANS:
Source of Variation
Treatments
Error
Total
PTS: 1
SS
47,511
41,400
88,911
df
2
15
17
MS
23,756
2,760
F
8.61
P-value
0.003
F critical
2.70
REF: SECTION 14.1
68. {GMAT Scores Narrative} Can he infer at the 10% significance level that the GMAT scores differ
among the three schools?
ANS:
H0: 1 = 2 = 3 vs. H1: At least two means differ
Conclusion: Reject the null hypothesis. Can say that the average GMAT score differs between the
three schools, according to this data.
PTS: 1
REF: SECTION 14.1
69. In a completely randomized design, 15 experimental units were assigned to each of four treatments.
Fill in the blanks (identified by asterisks) in the partial ANOVA table shown below.
Source of Variation
Treatments
Error
Total
SS
*
*
2512
df
*
*
*
MS
240
*
F
*
ANS:
Source of Variation
Treatments
Error
Total
SS
720
1792
2512
df
3
56
59
MS
240
32
F
7.5
PTS: 1
REF: SECTION 14.1
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copied, or distributed without the prior consent of the publisher.
70. In a completely randomized design, 12 experimental units were assigned to the first treatment, 15 units
to the second treatment, and 18 units to the third treatment. A partial ANOVA table is shown below:
Source of Variation
Treatments
Error
Total
a.
b.
SS
*
*
*
MS
*
35
F
9
Fill in the blanks (identified by asterisks) in the above ANOVA table.
Test at the 5% significance level to determine if differences exist among the three
treatment means.
ANS:
a.
Source of Variation
Treatments
Error
Total
b.
df
*
*
*
SS
630
1470
2100
df
2
42
44
MS
315
35
F
9
H0: 1 = 2 = 3 vs. H1: At least two means differ
Rejection region: F > F0.05,2,42  3.23
Test statistics: F = 9.0
Conclusion: Reject the null hypothesis. Yes, differences in means exist in at least two of
the three treatment means, according to this data.
PTS: 1
REF: SECTION 14.1
Mutual Funds
An investor studied the percentage rates of return of three different types of mutual funds. Random
samples of percentage rates of return for four periods were taken from each fund. The results appear in
the table below:
Mutual Funds Percentage Rates
Fund 1
Fund 2
Fund 3
12
4
9
15
8
3
13
6
5
14
5
7
17
4
4
71. {Mutual Funds Narrative} Set up the ANOVA Table. Use  = 0.05 to determine the critical value.
ANS:
Source of Variation
Treatments
Error
Total
PTS: 1
SS
252.40
49.20
301.60
df
2
12
14
MS
126.20
4.10
F
30.78
REF: SECTION 14.1
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copied, or distributed without the prior consent of the publisher.
72. {Mutual Funds Narrative} Test at the 5% significance level to determine whether the mean percentage
rates for the three funds differ.
ANS:
H0: 1 = 2 = 3 vs. H1: At least two means differ
Rejection region: F > F0.05,2,12 = 3.89
Conclusion: Reject the null hypothesis. Yes, the mean percentage rates differ for at least two of the
three mutual funds, according to this data.
PTS: 1
REF: SECTION 14.1
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copied, or distributed without the prior consent of the publisher.